3

Find this limit

$$\lim_{n\to\infty}n^2\left(\dfrac{1^k+2^k+\cdots+n^k}{n^{k+1}}-\dfrac{1}{k+1}-\dfrac{1}{2n}\right)\tag{1}$$

I can only solve this limit

$$I=\lim_{n\to\infty}\left(\dfrac{1^k+2^k+\cdots+n^k}{n^k}-\dfrac{n}{k+1}\right)=\dfrac{1}{2}$$ proof: use Stolz therom:let $$a_{n}=(k+1)(1^k+2^k+\cdots+n^k)-n^{k+1},b_{n}=(k+1)n^k$$ then $$I=\lim_{n\to\infty}\dfrac{a_{n}}{b_{n}}=\lim_{n\to\infty}\dfrac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}=\dfrac{1}{k+1}\dfrac{(k+1)(n+1)^k-(n+1)^{k+1}+n^{k+1}}{(n+1)^k-n^k}$$ so $$I=\lim_{n\to\infty}\dfrac{1}{k+1}\dfrac{[k(k+1)/2]n^{k-1}+\cdots+k}{kn^{k-1}+\cdots+1}=\dfrac{1}{2}$$ But for (1),I can't,Thanks.

Did
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math110
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  • Please do not modify significantly the requirements you put on a solution, nearly one year after answers got posted. (I seem to remember you already did that on other questions of yours, you just have to stop.) – Did Oct 21 '14 at 16:19
  • Faulhaber's formula, proved at this MSE link produces $kB_2/2 = k/12.$ – Marko Riedel Oct 28 '14 at 01:55

2 Answers2

5

For every $k\geqslant0$ and $n\geqslant0$, let $$ s_n^k=\sum\limits_{i=1}^ni^k. $$ We first recall how to get a simple equivalent of $s_n^k$ when $n\to\infty$. One starts from the fact that, for every $i\geqslant1$, $(i-1)^k\leqslant t^k\leqslant i^k$ for every $t$ in $(i-1,i)$. Summing this from $i=1$ to $i=n$ yields $$ (k+1)s_{n-1}^k\leqslant (k+1)\int_0^nt^k\mathrm dt\leqslant(k+1)s_n^k, $$ that is, $$ n^{k+1}\leqslant (k+1)s_n^k=(k+1)s_{n-1}^k+(k+1)n^k\leqslant n^{k+1}+(k+1)n^k, $$ which is enough to get $$ s_n^k=\frac{n^{k+1}}{k+1}+O(n^k)\tag{$\dagger$}. $$ To be more precise than $(\dagger)$, one needs to refine the inequalities $(i-1)^k\leqslant t^k\leqslant i^k$. For every $i\geqslant1$ and every $t$ in $(i-1,i)$, Taylor formula reads $$ t^k=i^k+ki^{k-1}(t-i)+\frac12k(k-1)r^{k-2}(t-i)^2, $$ for some $r$ in $(t,i)$. If $k\geqslant2$, $r^{k-2}\leqslant i^{k-2}$. Furthermore, for every $i\geqslant1$, $\int\limits_{i-1}^i(t-i)\mathrm dt=-\frac12$ and $\int\limits_{i-1}^i(t-i)^2\mathrm dt=\frac13$ hence, integrating both sides from $t=i-1$ to $t=i$ and summing over $i$ from $i=1$ to $i=n$ yields $$ n^{k+1}=(k+1)s_n^k-\frac12(k+1)ks_n^{k-1}+\frac16(k+1)k(k-1)s_n^{k-2}\theta_n^k, $$ where $\theta_n^k$ is in $(0,1)$. Since $ks_n^{k-1}=n^k+O(n^{k-1})$ and $s_n^{k-2}=O(n^{k-1})$, this guarantees that $$ s_n^k=\frac{n^{k+1}}{k+1}+\frac12n^k+O(n^{k-1}).\tag{$\ddagger$} $$ The expansion $(\ddagger)$ refines $(\dagger)$ but we need one more term. To get it, use $(\ddagger)$ for $k-1$. This reads $$ ks_n^{k-1}=n^{k}+\frac12kn^{k-1}+O(n^{k-2}), $$ Additionally, note that $s_n^{k-1}\leqslant s_n^{k-2}\theta_n^k\leqslant s_n^{k-2}$ hence $\theta_n^k\to1$ when $n\to\infty$.; Since $(k-1)s_n^{k-2}n^{k-1}+O(n^{k-2})$, putting everything together yields $$ n^{k+1}=(k+1)s_n^k-\frac12(k+1)\left(n^{k}+\frac12kn^{k-1}\right)+\frac16(k+1)kn^{k-1}+O(n^{k-2}), $$ that is, $$ \frac{n^{k+1}}{k+1}=s_n^k-\frac12n^{k}-\frac1{12}kn^{k-1}+O(n^{k-2}), $$ which is equivalent to $$ n^2\left(\frac{s_n^k}{n^{k+1}}-\frac1{k+1}-\frac1{2n}\right)=\frac1{12}k+O\left(\frac1n\right). $$

Did
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  • But the correct result should be $\frac 1{12}k$, I guess. – Hagen von Eitzen Nov 03 '13 at 15:58
  • @HagenvonEitzen Thanks for notifying me. Indeed the result is $\frac1{12}k$, as shown when one does not neglect a term of order $n^{k-1}$ as I foolishly did... – Did Nov 03 '13 at 16:51
2

We know that $\sum_{i=1}^n i^k=s_k(n)$ for some polynomial $s_k(x)=a_0+a_1x+\ldots +a_{k+1}x^{k+1}$. (We know this because the map $\Delta\colon f\mapsto f(X)-f(X-1)$ is a linear map from the $\mathbb Q$-vector space of polynomials of degree $\le k+1$ to the vector space of polynomials of degree $\le k$ and the kernel consists precisely of the onedimensional space of constant polynomials, hence $\Delta$ is onto). We want to take the limit of $$ \begin{align}&n^2\left(\frac{p_k(n)}{n^{k+1}}-\frac{1}{k+1}-\frac1{2n}\right)\\=&n^2\left[\left(a_{k+1}-\frac1{k+1}\right)+\left(a_k-\frac12\right)\frac1n+a_{k-1}\frac1{n^2}+O\left(\frac1{n^3}\right)\right]\\ \end{align}$$ and observe that the result is $a_{k-1}$ (provided that we can confirm that $a_{k+1}=\frac1{k+1}$ and $a_k=\frac12$). One can determine the highest $a_j$ step by step by comparing coefficients: $$\begin{align}n^k &=s_k(n)-s_k(n-1)\\ &=a_{k+1}(n^{k+1}-(n-1)^{k+1})+a_k(n^k-(n-1)^k)+a_{k-1}(n^{k-1}-(n-1)^{k-1})+\ldots \\ &=\hphantom{+\;}a_{k+1}\left((k+1)n^k-\tfrac{(k+1)k}{2}n^{k-1}+\tfrac{(k+1)k(k-1)}{6}n^{k-2}+O(n^{k-3})\right)\\ &\quad+\;a_{k}\left(kn^{k-1}-\tfrac{k(k-1)}{2}n^{k-2}+O(n^{k-3})\right)\\ &\quad+\;a_{k-1}\left((k-1)n^{k-2}+O(n^{k-3})\right)\\ &\quad +\;O(n^{k-3})\\ &=\hphantom{+\;}a_{k+1}(k+1)\cdot n^k\\ &\quad+\;\left(a_kk-a_{k+1}\tfrac{(k+1)k}{2}\right)\cdot n^{k-1}\\&\quad+\;\left(a_{k-1}(k-1)-a_k\tfrac{k(k-1)}{2}+a_{k+1}\tfrac{(k+1)k(k-1)}{6}\right)\cdot n^{k-2}\\&\quad +\;O(n^{k-3}),\end{align}$$ that is $$\begin{align}a_{k+1}&=\frac1{k+1}\\ a_k&=\frac1k\cdot a_{k+1}\frac{(k+1)k}{2}=\frac12\\ a_{k-1}&=\frac1{k-1}\left(a_k\frac{k(k-1)}{2}-a_{k+1}\frac{(k+1)k(k-1)}{6}\right)=\frac k{12}.\end{align} $$ Thus the final result is $$ \lim_{n\to\infty}n^2\left(\frac{1^k+2^k+\ldots+n^k}{n^{k+1}}-\frac{1}{k+1}-\frac1{2n}\right)=\frac k{12}.$$

Hagen von Eitzen
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